Maths for Construction - Desklib

This document contains solved problems on Maths for Construction including scenarios on equations, replacement policies, hypothesis testing, and wave displacement. The solutions are explained step by step with relevant formulas and assumptions.

MATHS FOR CONSTRUCTION
Name of Student
Institution Affiliation

Task 1
Scenario 1
A=LW; L=w+3.2
26.5= (w+3.2) w; w2+3.2w-26.5=0
w=−3.2 ± √3.22−(4 × 1×−26.5)
2
=4.4 or -10.8
(b)
Lift =k × ρ ×V 2 × A
L = (1/2) d v2 s CL
The angle of attack and CL are related and can be found using a Velocity Relationship Curve
Graph
Scenario 2
a)
65 miles → 60 seconds
65 miles= (65 ×1760 /0.91) m=104104 m
Speed in m/s=104104/60=1735 m/s
65 miles=1 hour; 100 miles=10/65=1.54 hours

30 miles/gallon
1 gallon→3.78 liters
1 mile →1.61 km
30 miles →(30*1.61) km
Consumption in litters/km
=3.78/ (30*1.61)=0.078liters/km
Fuel consumption for the whole journey
{ 3.78
30 ×1.61 × 100
1.61 }=4.844
b)
L = (1/2) d v2 s CL
L = Lift, which must equal the airplane's weight in pounds
d = density of the air. This will change due to altitude. These values can be found in
a I.C.A.O. Standard Atmosphere Table.
v = velocity of an aircraft expressed in feet per second
s = the wing area of an aircraft in square feet
CL = Coefficient of lift, which is determined by the type of airfoil and angle of attack.
Scenario 3
1)
6th term

nth term=a+ ( n−1 ) d
d=2/3b-b=-1/3b
6th term=b+ (6-1)*-1/3b
=-2/3b
kth term
kth term=a+ (k-1)*-1/3b
20th term=15; 20th term=b-19/3b=-16/3b
−16
3 b=15 ; b=−45
16
Sum of the first 20 terms
Sn= n
2 ( a+ an ) = 20
2 (−45
16 +15)
=121.875
nth term=arn−1
2)
20th term=1 × 1
2
19
= 1
2
19
Sum to infinity
Sn= a .1(1−rn )
1−r

lim
n → ∞
a(1−rn)
1−r
Since |r|<1, then rn
→ 0 as n → ∞
Hence
S∞ = a
1−r = 1
1−1
2
=2
2Log (3x) + Log (18x) = 27
log (3x) 2+log 18x=27
9x2+18x=27
X2+2x-3=0
x=−2 ± √22− ( 4 ×1 ×−3 )
2 =1∨−3
3)
2 LOGe (3x) + LOGe (18x) = 9
LOGe (3x) 2+LOGe (18x) = 9
3x2+18x=9
9x2+18x-9=0
X2+2x-1=0

¿−2± √ 22 − ( 4 × 1×−1 )
2 =0.414∨−2.414
Cosh(X) + Sinh(X) = 5
ex+e−x
2 +( ex−e−x
2 )=9
ex+e−x−ex+ e−x
2 =9; 2 ex
2 =9
ex=9
Let ex be p
p=9
x=ln 9=2.19
Cosh (2Y) - Sinh (2Y) = 3
Cosh (2Y) = 1
2 ( e2 y+ e−2 y ) ; Sinh (2Y) = 1
2 ( e2 y−e−2 y )
( e2 y +e−2 y )− ( e2 y−e−2 y ) =3
( e2 y +e−2 y )− ( e2 y +e−2 y )=3
2 e−2 y−3=0 ; e−2 y= 2
3
2 y=ln 2
3 → y= 1
2 ln 2
3

Cosh (K) * Sinh (K) = 2
Cosh (K) = 1
2 ( e2 y+ e−2 y ) ; Sinh (K) = 1
2 ( e2 y−e−2 y )
( e2 y +e−2 y )− ( e2 y−e−2 y ) =2
( e2 y +e−2 y )− ( e2 y +e−2 y )=2
2 e−2 y−3=0 ; e−2 y= 2
3
2 y=ln 2
3 → y= 1
2 ln 2
3
Cosh (M) / Sinh (M) = 2
Cosh (M) = 1
2 ( e2 y+ e−2 y ) ; Sinh (M) = 1
2 ( e2 y−e−2 y )
( e2 y +e−2 y )− ( e2 y−e−2 y ) =2
( e2 y +e−2 y )− ( e2 y +e−2 y )=2
2 e−2 y−3=0 ; e−2 y= 2
3
2 y=ln 2
3 → y= 1
2 ln 2
3
Task 2
Scenario 1

0-4 5-9 10-14 15-19 20-29 30-39
0
5
10
15
20
25
30
35
40
45
0 2 4 6 8 10 12
0
50
100
150
200
250
Revenue
Customers

0-4 5-9 10-14 15-19 20-29 30-39
0
10
20
30
40
50
60
70
80
0 2 4 6 8 10 12
0
2
4
6
8
10
12
Revenue
Customers
(c)
Mean=∑ fx/ ∑ f
=1630.5/144=11.323
Range=5 for the first 4 groups and 1o for the last two groups
Variance σ2 = ∑ fx2
∑ f −x2
= (26652.25/144)-11.3232

56.875
Standard deviation=7.54
Scenario 2
Kolmogorov-Smirnov (K-S) and the Shapiro-Wilk (S-W) tests can be used in the testing of the
assumption that the data sample are extracted from a normally distributed population.
The probabilities given in the problem are cumulative i.e. till week 1, till week 2 etc. Individual
probabilities would be 0.10 in 1st week, 0.15 (0.25-0.10) in 2nd week, and so on.
Individual Failures/week = Total Quantity / Mean Life = 1000 / 3.45 = 289.9
Individual Replacement Cost = (Individual Failures per week) x (Individual replacement cost)
= 289.9 x 3 = Rs. 869.6
In the first week: 10 % (0.10) of the bulbs will fail out of 1000 bulbs i.e. 100
In the second week: 15 % of the bulbs will fail out of 1000 bulbs i.e. 150. Also, 10% of 100
replaced in the first week i.e. 10. TOTAL bulbs failed until second week = 160 (150+10)
Thus, replacing all the bulbs simultaneously at fixed interval & also to replace the individual
bulbs that fail in between will be economical or optimal after 4 weeks (optimal interval between
group replacements).
Interpretation:
1. The cost of only individual replacements id Rs. 869.6
2. The cost of combine policy i.e. group and individual replacement id Rs. 863.6
3. Hence the Policy-II is the optimum replacement policy
Hence, the bulbs shall be replaced every four weeks individually as well as in groups which
combine would cost Rs. 863.6 per week

[Note: "Yes we can, if..." A way to help solve this type of problem is to answer "Yes we can,
if..." In this case the question is, "Can it be concluded that the mean age of the population is not
30?" Answer, "Yes we can, if we can reject the null hypothesis that it is 30." Responding to
problems the same way all the time will lead to less confusion and less errors. ]
(1) Data
n = 10 = 20
= 27 = .05
(2) Assumptions
 simple random sample
 normally distributed population
(3) Hypotheses
: = 30
: 30
(4) Test statistic
as the population variance is known, we use z as the test statistic.
(a) Distribution of test statistic
if the assumption is correct and is true, the test statistic follows the standard normal distribution.
Therefore, we calculate a z score and use it to test the hypothesis.
(b) Decision rule
Reject if the z value falls in the rejection region. Fail to reject if it falls in the non-
rejection region.

Because of the structure of it is a two tail test. Therefore, reject if z -1.96 or z 1.96.
(5) Calculation of test statistic
(6) Statistical decision
the null hypothesis is rejected because z = -2.12 which is in the rejection region. The value is
significant at the .05 level.
(7) Conclusion
it is concluded that is not 30.
p = .0340
A z value of -2.12 corresponds to an area of .0170. Since there are two parts to the rejection
region in a two tail test, the p value is twice this which is .0340.
A problem like this can also be solved using a confidence interval. A confidence interval will
show that the calculated value of z does not fall within the boundaries of the interval. It will not,
however, give a probability.

Confidence interval
Same example as one tailed test
(1) Data
n = 10 = 20
= 27 = .05
(2) Assumptions
 simple random sample
 normally distributed population
(3) Hypotheses
: = 30
: 30
(4) Test statistic
as the population variance is known, we use z as the test statistic.
(a) Distribution of test statistic
if the assumptions are correct and are true, the test statistic follows the standard normal
distribution. Therefore, a z score is calculated and used to test the hypothesis.
(b) Decision rule

Reject if the z value falls in the rejection region. Fail to reject if it falls in the non-
rejection region.
With = .05 and the inequality we have the entire rejection region at the left. The critical value
will be z = -1.645. Reject if z < -1.645.
(5) Calculation of test statistic
(6) Statistical decision
the null hypothesis is rejected because -2.12 < -1.645.
(7) Conclusion
it is concluded that < 30.
p = .0170 this time because it is only a one tail test and not a two tail test.
Task 3
In the equation x=sin (wt+Ɵ), the amplitude is A, phase is Ɵ, the frequency is f=w/2 π and the
period T=1/f=2 π/w
A=3.75
Phase=2 π/9

Frequency=100 π /2 π=50
Periodic time=1/f=1/50=0.02 seconds
Maximum wave displacement occurs when the argument is π /2 (maximum positive
displacement) or 3 π /2 (maximum negative displacement). Having each of the arguments set at
π / 2
3.75 sin (100𝜋𝑡 + 2𝜋/9) =π /2
sin (100𝜋𝑡 + 2𝜋/9) =π /2/3.75
100𝜋𝑡 + 2𝜋/9=24.8; 100𝜋𝑡=24.102
t=0.706 s
4.42 sin (100𝜋𝑡 − 2𝜋/5) =π /2
sin (100𝜋𝑡 − 2𝜋/5) =π /2/4.42
100𝜋𝑡 − 2𝜋/5=20.8; 100𝜋𝑡=22.1
t=0.646 s
vi.
t 0 0.002 0.00
4
0.00
6
0.00
8
0.01 0.01
2
0.01
4
0.01
6
0.01
8
0.02
x1 0.046 0.087 0.12
8
0.16
9
0.21 0.25
1
0.29
2
0.33
3
0.37
4
0.41
5
0.456
x2 -0.097 -0.048 0 0.04
8
0.09
7
0.14
5
0.19
4
0.24
2
0.29
1
0.33
9
0.387
x1+x -0.051 0.039 0.12 0.21 0.30 0.39 0.48 0.57 0.66 0.75 0.843

2 8 7 7 6 6 2 5 4
vii.
0 0.005 0.01 0.015 0.02 0.025
-0.2
0
0.2
0.4
0.6
0.8
1
x1
x2
x1+x2
t (s)
Ɵ (⁰)
Scenario 2
Distance AB
√( 40−0)2 +(0+40)2 +(−20−0)2
√3600=60 units
Angle between AB and BC
(0, -40, 0); (3, 4, 1)
(0*3)+ (-40*4) + (0*1)
=-160

||u||= √ u1
2 +u2
2=√ 02 +−402+02=40
||v||= √v1
2 + v2
2=√32 +42+ 12=3 √ 3
cos θ u . v
||u||.∨|v|∨¿ = 160
40× 3 √3 ;θ=cos−1 0.7698 ¿
θ=39.7 °
Vector equation
(a
b
0 )−
( 40
0
−20)=
(a−40
b
20 )
r =
( 40
0
−20 ) +ℷ ( a−40
b
20 )
x=40+ℷ ( a+ 40 ) , y=− bℷ , z=−20+ ℷ 20
ℷ= x−40
a+ 40 , ℷ= y −0
b , ℷ= z+ 20
20
x−40
a+ 40 = y−0
b = z +20
20

Task 4
Scenario 1
(a)
-60 -40 -20 0 20 40 60 80
-20000
0
20000
40000
60000
80000
100000
120000
x
M
The range of values where the above Bending Moment Function is maximum or minimum,
decrease or increasing
M=3000-550x-20x2
When M=0
x=−550 ± √ 5502−(4 ×20 ×−3000)
40 ; 4.66∨−32.1625

dM/dx=40x+550
When x=4.66
x -4 4.66 8
dM/dx 390 736.4 870
When x=-32.1625
x -60 -32.1625 10
dM/dx -1850 -736.5 950
From the calculations, the function is decreasing
The temperature, Ɵ (⁰C) at time t (mins) of a body is given by
Ɵ=300 +100e-0.1t
Evaluate Ɵ for t=0, 1, 2 and 5
Ɵ=300 +100e-0.1(0) =400
Ɵ=300 +100e-0.1(1) =390.48
Ɵ=300 +100e-0.1(2) =381.87
Ɵ=300 +100e-0.1(5) =360.65
Range of temperature for positive t= (360.65 to 400)
(c) 0=300 +100e-0.1t
100e-0.1t=-300; e-0.1t=-3
0.9048t=-3; t=-3.3
log (P) +n log (V)-log C

log (P) + log (V) n=log C
P.Vn=C; hence
PVn=C
log (P) +n log (V)-log C
log (P) +n log (V) =C
log 10+n log V=1; 2 log V=1
log V=0; log V=-2
V=0.01
log 60+ 2 log V=1
1.778+2 log V=1; 2 log V=-0.778
log V=-0.389
V=0.408
Scenario 2
(a)

0 1 2 3 4 5 6 7 8 9
0
200
400
600
800
1000
1200
(b) Calculus can be used in the determination of the time for maximum and minimum production
and the time at which there will be no production at all. Maximum production occurs when C is
maximum and the reverse is true for minimum production. Similarly calculus is usable in the
determination of the amount of production at various production times hence allowing a
prediction of the trend of production.
(c) C=16t-2+2t-
0=16t-2+2t-; 16t-2+2t-=0
16t-2=-2t-
t-1=-1/8; t=-8
(d) dC/dt=32t-3-2t-2
t -20 -8 4
dC/dt -0.009 -0.09375 0.375
dC/dt =32t-3-2t-2

32(-20)-3-2(-20)-2=-0.004-0.005
=-0.009
dC/dt=32t-3-2t-2
32(-8)-3-2(-8)-2=-0.0625-0.03125
=-0.09375
dC/dt=32t-3-2t-2
32(4)-3-2(4)-2=0.5-0.125
=-0.375
There would no more minimum cost of production since the graph turns only at one point which
is at 8 seconds. This would turn out as the only minimum production time and thus minimum
production cost.
Scenario 3
-25 -20 -15 -10 -5 0 5 10 15 20 25
0
500000
1000000
1500000
2000000
2500000
3000000
3500000
E3t
t